内容正文:
高三数学(文科) 答案 第 1 页 共 9 页
2018 届高三模拟考试
高三数学(文科)参考答案及评分标准 2018.3
一、选择题:本大题共 12 小题,每小题 5 分,共 60 分.
ACBD ABCB ADAD
二、填空题:本大题共 4 小题,每小题 5 分,共 20 分.
13. 2 14. 3 15. 2 2( 2) 2x y+ − = 16.
3
[
4
,
7
]
8
三、解答题:本大题共 6 小题,共 70 分.解答应写出文字说明、证明过程或演算步骤.
17.(I)解: 1 1 4.a S= = ······························································································· 2 分
当 2n� 时, 1
2 23 5 3( 1) 5( 1)
3 1.
2 2n n n
n
S
n
a S
n n
n−
+ − + −
− == +− = ····················· 6 分
又 1 4a = 符合 2n� 时 na 的形式,所以{ }na 的通项公式为 3 1.na n= + ················ 8 分
(II)由(I)知 nb =
3 1 1
.
(3 1)(3 4) 3 1 3 4n n n n
= −
+ + + +
················································ 10 分
数列{ }nb 的前n项和为
1 2
1 1 1 1 1 1 1 1
( ) ( ) ( ) ( )
4 7 7 10 3 2 3 1 3 1 3 4n
b b b
n n n n
+ + + = − + − + + − + −
− + + +
L L
1 1
4 3 4n
= −
+
. ············································································· 12 分
18. (I)证明:由底面 ABCD 为矩形,得 BC ⊥ .AB
又平面SAB ⊥ 平面 ABCD ,平面SAB ∩平面 ABCD AB= , BC ⊂ 平面 ABCD ,
所以 BC ⊥ 平面 .SAB 所以 .BC SA⊥ ···································································· 3 分
同理可得 .CD SA⊥ ······························································································· 5 分
又 BC CD C=∩ , BC ⊂ 平面 ABCD ,CD ⊂平面 ABCD ,
所以SA ⊥ 平面 .ABCD …………………………6 分
(II)解:设 6SA a= ,则 2AB a= , 3 .AD a=
3
1
3
1 1 1
= ( ) ( )
3 2 2
1 1
( 2 3 ) (3 ) 3 .
3 2
E BCD BCDV S h
BC CD SA
a a a a
− = × ×
× × × ×
= × × × × =
△
B
E
DA
C
S
高三数学(文科) 答案 第 2 页 共 9 页
又
8
9E BCD
V − = ,所以
3 83 .
9
a = 解得
2
.
3
a = ····························································· 9 分
四棱锥S ABCD− 的外接球是以 AB、AD 、AS 为棱的长方体的外接球,设半径为 .R
则
2 2 2 142 7
3
R AB AD AS a= + + = = ,即
7
.
3
R = ···············································11 分
所以,四棱锥S ABCD− 的外接球的表面积为 2
196π
4π .
9
R = ····························· 12 分
19. 解:(Ⅰ)由样本估计总体的思想,甲高中学生一周内平均每天学习数学的时间的中位
数
0.5 (0.1 0.2)
=20 10 26.67
0.3
m
− +
+ × ≈
甲
; ···